BEE1020 — Basic Mathematical Economics Dieter Balkenborg
January 2004 exam — Solutions
Department of Economics
02.02.2004
University of Exeter
Section A
(Basic algebra and calculus – multiple choice)
Question A1 : The function f (x) =
(
(x−2)x2 (x+2)
(x+2)(x−2)
4
for x > 0 and x 6= 2
is
for
x=2
h1i not well-defined.
h2i convex.
h3i concave.
h4i discontinuous.
h5i not differentiable.
h6i decreasing.
Solution A1 : h2i
Question A2 : Which of the following functions is non-negative on the interval −2 <
x < 2:
h1i y = x3 (x3 − 1)
h2i y = x2 − 2x + 1
h3i y = x2 (x2 − 4)
h4i y = x (x − 4) (x + 4)
h5i y = x4 (x4 − 1)
h6i y = 4x3 − 3x2 + 2x − 1 convex.
Solution A2 : h2i
Question A3 : The inequality (x − 2)2 (x − 1) x2 (x + 1) (x + 2) ≤ 0 can be solved as
h1i x ≤ −2 or −1 ≤ x ≤ 0 or 1 ≤ x ≤ 2
h2i −2 ≤ x ≤ 1 and x = 0 and 1 ≤ x ≤ 2
h3i −2 < x < 1 and x = 0 and 1 < x < 2
h4i −2 < x < 1 or 0 ≤ x < 2
h5i x ≤ −2 and −1 ≤ x ≤ 0 and 0 ≤ x ≤ 1 and x = 2
h6i x ≤ −2 or −1 ≤ x ≤ 1 or x = 2
Solution A3 : h6i
Question A4 : A consumer has spent $1,000 on designer clothes. Current VAT is
17.5%. How much VAT did the consumer pay.
h1i $175
h2i $17.50
h3i $851.49
h4i $148.94
h5i$0.00
h6i Solution cannot be determined because of insufficient information.
Solution A4 : h4i
Question A5 : Which statement is correct for the function f (x) = xn with n an
integer?
h1i If n is odd and positive then x = 0 is a peak.
h2i If n is odd then the function is concave on the whole real line.
h3i If n is odd and negative then the function is concave on the negative numbers.
h4i If n is even then the function is convex on the whole real line.
h5i If n is even and negative then the function is both positive- and negative-valued.
h6i If n is even and positive then the function is non-decreasing.
Solution A5 : h3i
2
Question A6 : You are given the following information about the signs of the first and
second derivative of a function y (x):
x
2 4 7 11
0
y (x) − + + −
y 00 (x) + + − +
Which of the following six graphs is consistent with this information?
100
2
0
-100
4
x6
8
10
12
-200
-300
-400
200
200
180
180
160
160
140
140
120
120
100
100
80
80
60
60
40
40
20
20
-500
0
2
Graph 1
4
x6
8
10
12
150
100
50
4
6
x
8
10
12
200
200
180
180
160
160
140
140
120
120
100
100
80
80
60
60
40
40
0
Graph 4
4
6x
8
10
12
10
12
Graph 3
20
2
2
Graph 2
200
0
0
20
2
4
6x
8
10
12
0
2
Graph 5
4
6x
Graph 6
Solution A6 : h3i
Question A7 : The following statement is false for the function f (x) = x4 :
h1i On the interval −2 ≤ x ≤ 3 the function has a minimum at x = 0.
h2i The function has a unique maximum on the interval −2 ≤ x ≤ 3.
h3i The function is strictly convex.
h4i The function has an inflection point at x = 0.
h5i The function is decreasing on the non-negative numbers.
h6i The function has a unique root.
Solution A7 : h4i
3
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Section B
(Basic algebra and calculus – exercise)
Question B1 : Simplify (x − 2) (x9 − x + 3) (x + 2) − (x − 2) (x9 + x + 3) (x + 2)
Solution B1 : −2x (x − 2) (x + 2) = 8x − 2x3
2
Question B2 : Simplify
(xy2 )3
2
Solution B2 :
3
(5x2 y3 ) −(10x3 y2 )
3
(5x2 y3 ) −(10x3 y2 )
(xy 2 )3
=
25x4 y 6 −1000x9 y6
x3 y6
= 25x − 1000x6 = 25x (1 − 40x5 )
Question B3 : Consider the function y = x2 +4x+2 on the domain {x ≥ 0}. Draw the
graph of the function, Is the function invertible. If yes, what is the inverse function?
Solution B3 :
40
30
20
10
0
1
2
x
3
4
5
2
The function
is invertible.
√
√ The inverse is given by x + 4x + 2 − y = 0, x =
−2 + 4 − 2 + y = −2 + 2 + y. (We must take the larger root.)
Question B4 : Find all solutions to the quadratic equations
i) 4x2 + 4x = −2
ii) 4x2 + 4x = 0
iii) 4x2 + 4x = 8
q
q
1
1
1
1
Solution B4 : i) x = − 2 ± 4 − 2 (no real solution) ii) x = − 2 ± 14 = 0 or −1 iii)
q
q
1
1
1
x = − 2 ± 4 + 2 = − 2 ± 94 = 1 or −2
Question B5 : Solve the simultaneous systems of equations
i) 8 = 2x + 6y, 9y = 12 − 6x
ii) 8 = 4x + 6y, 9y = 12 − 6x
Solution B5 : i) x = 0, y = 1 13 ii) x arbitrary, y =
4
3
− 23 x
Question B6 : Solve the simultaneous system of equations x2 + y 2 = 5, y + 2x = 0.
Solution B6 : x2 + 4x2 = 5x2 = 5, x = ±1, y = ∓2.
Question B7 : Use a sign diagram to find all solutions to the inequality (x + 1)2 (x2 − 4) <
0.
4
Solution B7 : y = (x + 1)2 (x2 − 4) = (x + 1) (x + 2) (x − 2)
x < −2 x = −2 −2 < x < −1 x = −1 −1 < x < 2 2 2 < x
x+2
−
0
+
+
+ +
+
2
(x + 1)
+
+
+
0
+ +
+
x−2
−
−
−
−
− 0
+
y
+
0
−
0
− 0
+
Thus y < 0 if −2 < x < −1 or −1 < x < 2.
Question B8 : Factorize the polynomial y = 4x3 − 7x + 3
Solution B8 : x = 1 is a root
4x2 +4x −3
x −1 | 4x3
−7x +3
4x3 −4x2
4x2 −7x +3
4x2 −4x
−3x +3
q
4x2 +4x−3 = 0 has the solutions x = − 12 ± 14 + 34 = 12 or − 32 . This quadratic poly¡
¢¡
¢
nomial has hence the factorizations 4 x − 12 x + 32 = (2x + 3) (2x − 1). Overall
y = 4x3 − 7x + 3 = (x − 1) (2x + 3) (2x − 1)
Question B9 : Consider the first, second, third and so on derivative f (n) (x) of the
polynomial x4 + x. From which n onwards will all derivatives f (n) (x) be constant?
Solution B9 : f 0 = 4x3 + 1, f 00 = 12x2 , f 000 = 24x, f (iv) = 24, f (v) = f (vi) = . . . = 0.
The fourth and all higher derivatives are constant.
Question B10 : What is the slope of the tangent to the graph of the function
¡
¢3
y = x3 − x2 + x − 1
at x = 2?
Solution B10 :
¡
¢2 ¡
¢
y 0 = 3 x3 − x2 + x − 1 3x2 − 2x + 1
y 0 (2) = 3 (8 − 4 + 2 − 1)2 (12 − 4 + 1) = 3 × 52 × 9 = 675
Question B11 : Is x = 0 an inflection point of the function y = x5 + x3 ?
Solution B11 : Yes the second derivative y 00 = 20x3 + 6x = (20x2 + 6) x changes sign
at 0.
5
Question B12 : Differentiate the function
y=
x2 + x + 1
x2 − x + 1
You can obtain one additional mark for simplifying.
Solution B12 :
y0 =
Section C
(2x + 1) (x2 − x + 1) − (x2 + x + 1) (2x − 1)
−2x2 + 2
=
(x2 − x + 1)2
(x2 − x + 1)2
(Applied calculus)
Question C1 : A producer operating in a perfectly competitive market has the total
cost function
T C (Q) = 2Q3 − 12Q2 + 30Q + 50
where costs are given in Pounds Sterling.
C1-i :
Calculate and sketch the marginal cost function MC (Q) and the average
variable cost function AV C (Q). (1 mark)
Solution C1-i: MC (Q) = 6Q2 − 24Q + 30, AV C (Q) = 2Q2 − 12Q + 30.
50
40
30
20
10
0
1
2
Q
3
4
5
C1-ii : Solve the equation MC (Q) = AV C (Q). (1 mark)
Solution C1-ii:
6Q2 − 24Q + 30 = 2Q2 − 12Q + 30 ⇔ 4Q2 − 12Q = 4Q (Q − 3) = 0
⇔ Q = 0 or Q = 3
C1-iii : At what quantity are average variable costs minimized? What are the minimum
average variable costs? (2 marks)
6
Solution C1-iii: Average costs are minimized at Q = 3, where AVC curve intersects
MC. AV C (3) = 2 × 9 − 4 × 9 + 30 = 12.
C1-iv : What quantity maximizes profits when the market price is P = 10? What is
the maximal profit the firm can make at this price? (2 marks)
Solution C1-iv: Price is below minimal AVC, so Q = 0 is optimal
C1-v : What quantity maximizes profits when the market price is P = 20? (1 mark)
Solution C1-v: The profit maximizing quantity is the larger solution Q = 5 to MC (Q) =
2Q2 − 12Q + 30 = 20. Solution is: Q = 5.
Question C2 : A manufacturer can produce 40 wardrobes at the cost of $37, 000 and
70 wardrobes at the cost of $55, 000. Assuming a linear cost function, what are his
marginal costs and what are his fixed costs? (3 marks)
Solution C2: Building 30 wardobes more costs $18.000 more. The marginal costs are
hence $600. This gives a variable cost of $24,000 to build 40 wardrobes. The fixed costs
are hence $13,000.
Question C3 : A manufacturer thinks that the economy is in a slump and he has to
cut prices. He has been selling lamps at $7 apiece, and at this price consumers have been
buying 6,000 lamps per month. The manufacturer estimates that for each $1 reduction
in price, 6,000 additional lamps can be sold each month. The manufacturer can produce
at costs $2 per lamp. At what price should the manufacturer sell the lamps to generate
the greatest possible profit? (4 marks)
Solution C3:
Π (∆p) = (7 + ∆p − 2) (6000 − 6000∆p) = 6000 (5 + ∆p) (1 − ∆p)
roots at ∆p = −5 and ∆p = 1, optimal price increase in the middle at ∆p = −2. Thus
price reduction by $2 optimal.
7
Section D
(Economic Applications)
Question D1 :
√
D2-i : A price taking producer has the production function Q = L. The costs of the
labour input L are given by the wage rate w > 0. If the output price is normalized
to 1, what is the maximal profit Π (w) of the producer as a function of the wage
rate? (4 marks)
Solution D1-i: Π (w) = maxL π (L, w) = maxL
√
¡
¢
1
1
1
L = 2w
, Π (w) = π 4w1 2 , w = 2w
− w 4w1 2 = 4w
.
h√
i
L − wL , FOC: π 0 =
1
√
2 L
− w = 0,
D2-ii : Given Π (w) calculate for each L > 0 the minimal value G (L) = minw>0 Π (w)+
wL. What function do you obtain? Can you give an economic interpretation of the
result? (4 marks)
1
4w
L
√
2 L
Solution D1-ii: G (L) = minw>0 H (w, L) = minw>0 Π (w)
³ + wL
´ = minw>0
FOC:− 4w1 2
+ L = 0, w2 =
1
,
4L
w=
1
√
,
2 L
G (L) = minw>0 H
1
√
,L
2 L
=
1
4
1
√
2 L
−
+ wL,
√
= L
Question D2 : A monopolist faces the demand D (p) = 10 − p units of a commodity
at price p which costs him $2 per unit to produce. Determine without using calculus
the optimal profit and verify using only algebra that this is indeed the optimal profit. (4
marks)
Solution D2 : The profit Π (p) = (p − 2) (10 − p) is zero at p = 2 and p = 10. Therefore
it is maximized in the middle at p = 6. Indeed, Π (p) = −p2 + 12p − 20 = − (p − 6)2 +
36 − 20 = 16 − (p − 6)2 is maximized at p = 6 with profit 16.
(BEE1020 January 2004)
(End of solutions.)
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